Chemistry 125: Lecture 50 February 11, 2011 Electrophilic Addition with Nucleophilic Participation Cycloaddition Epoxides This For copyright notice see.

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Presentation transcript:

Chemistry 125: Lecture 50 February 11, 2011 Electrophilic Addition with Nucleophilic Participation Cycloaddition Epoxides This For copyright notice see final page of this file

Problem: Suggest a Multi-Step Mechanism for the Acid-Catalyzed “Pinacol Rearrangement” (draw nice curved arrows) H+H+ CH 3 CC OH CH 3 OH CH 3 CC O + H 2 O CH 3 CC + OH CH 3 CC + O H Methide Shift + Driving Force?

Other “Simultaneous” Reagents Cl 2 C: (Carbene) R 2 BH (Hydroboration) CH 2 I 2 Zn/Cu (Carbenoid) O 3 (Ozonolysis) H-metal (Catalytic Hydrogenation) R-metal (Metathesis, Polymerization) RC (Epoxidation) OOHOOH O

Simmons-Smith “Carbenoid” Metal R-X Metal + R-X single-electron transfer (SET) e Metal + R X Metal R-M X + Zn Cu “couple” CH 2 I 2 + CH 2 The next three slides suggest a plausible, but incorrect, two-step mechanism for addition of ICH 2 ZnI to H 2 C=CH 2

ClZnCH 3 Model for I-Zn-CH 2 I 4s Zn LUMO 4p Zn LUMO + 1 bent for transition state LUMO` 4sp n Zn HOMO

Model for I-Zn-CH 2 I LUMO`  Zn-C HOMO

Model for I-Zn-CH 2 I ClZnCH 3 CH 2 ClZn CH 3 IZn CH 2 I HOMO  Zn-C LUMO  C-I ZnI 2 CH 2 “S N 2” If it were the diiodide instead of the model… But these two transition states were just guessed, not calculated quantum mechanically…

Although the above two-step mechanism with intermediate IZnCH 2 -CH 2 -CH 2 I is plausible, addition of IZnCH 2 I to H 2 C=CH 2 probably occurs in a single step, according to quantum mechanical calculation*, with the bent transition state shown below: * A DFT Study of the Simmons-Smith Cyclopropanation Reaction. A. Bottoni, et al., J. Am. Chem. Soc, 1997, 119, 12300

ICH 2 ZnI Zn I I

ICH 2 ZnI (at Transition State Geometry) LUMO Mixes with  HOMO HOMO-2 Mixes with  * LUMO Zn I I

meta-chlorobenzoic acid meta-chloroperoxybenzoic acid Epoxidation by Peroxycarboxylic Acids J&F sec 10.4a mCPBA 25°C 81% yield R = n-hexyl benzene 5 hr ?

LUMO HOMO Orbital Energy (kcal/mole) UMOs OMOs etc. Peroxyformic Acid Distorted to Transition State for O Transfer p(  O  * O-O “  -allylic”

CH 2 H2CH2C H2CH2C O O O C H H “S N 2 at O” “  -allylic” resonance p(  ) O nucleophile (nearby)  * O-O electrophile All happen together with minimal atomic displacement carboxylate “leaving group” (but not strictly in parallel)  C-C nucleophile p C+ electrophile  * H-O+ electrophile “S N 2 at H” backside attack

Transition State Geometry O-O Strongly Stretched (from ~1.5Å) O-H Hardly Stretched (from ~1Å) k H /k D ~ 1 Coplanar “Butterfly” mechanism (not spiro) suggested by Paul D. Bartlett (1950) calculated J. Amer. Chem. Soc. (1991) pp downhill motion after TS Only one TS : “Concerted but not Synchronous” “spiro” means two perpendicular rings sharing a common atom (here O 1 )

Note that arrows were not used as carefully in those days. Bartlett 1950 Problem: How about now? (compare arrows in this textbook illustration with the mechanism on the previous frames and try drawing a more accurate diagram)

Stereospecificity of Epoxidation: Concerted Syn Addition Pasto & Cumbo 1965 ~0°C 10 hr H H C C H3CH3C CH 3 O >99.5% trans mCPBA H H C C H3CH3C CH 3 trans OO 52-60% yield OO mCPBA HH C C H3CH3C CH 3 cis >99.5% cis H H C C H3CH3C CH 3 O ~0°C 10 hr52-60% yield

Alternative Epoxide Preparation (1936) H 2 O < 0°C 3 hr H H C C H3CH3C CH 3 HOCl Wilson & Lucas 1936 H H C C H3CH3C CH 3 Cl + H H C C CH 3 Cl HO 55% yield (distilled) SN2SN2 H2OH2O H 2 O 90°C 2 hr KOH (20M) H H C C CH 3 Cl -O-O H H C C H3CH3C CH 3 O 90% yield 45% over two steps syn inversion 2 nd inversion

CH 2 C H O H C C H OH CH 2 H OR O OEt O O CO 2 Et Ti RO O Remember Sharpless Asymmetric Epoxidation R ROO RO + Ti O O O R OEt O CO 2 Et O RO Ti O O O OEt O CO 2 Et ** R CH 2 H C C C H H allyl alcohol (R)-“epoxide”  (S)-epoxide precursor Chiral “Oxidizing Agent” LUMO? HOMO? is diastereomeric! ( also p O +  * C=C ) Cf. J&F Sec. 10.4b p. 426

20,000,000 tons $20 billion per year OLD CAMPUS H2CH2CCH 2 O H 2 C=CH 2 + O 2 Ag 250°C 15 atm ethylene oxide (84%) Raising the yield by 5% would be worth >$10 9 /year. * * The rest oxidizes to CO 2 /H 2 O. Only 0.05% of ethylene oxide is used as such.

H + Catalysis H2CH2CCH 2 O 20,000,000 tons $20 billion per year H2CH2CCH 2 O ethylene glycol (antifreeze, solvents, polymers) J&F Sec. 10.4c pp H2CH2CCH 2 HO OH H2OH2O of which 2/3 H2CH2CCH 2 O HO - Catalysis

End of Lecture 50 February 11, 2011 Copyright © J. M. McBride Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0